Question Detail
Question
A body of radius R and mass m is rolling smoothly with speed v on a horizontal surface. It then rolls up a hill to a maximum height h = 3v²/4g, what is the body's rotational inertia about the rotational axis through its centre of mass?
Correct Answer
Option A is the correct answer.
Detailed Explanation
Conservation of energy: ½mv² + ½Iω² = mgh. For rolling without slipping, ω = v/R. So ½mv² + ½I(v²/R²) = mg(3v²/4g) = 3mv²/4. Multiply by 2: mv² + I(v²/R²) = 3mv²/2. So I(v²/R²) = 3mv²/2 - mv² = mv²/2. Thus I = (mR²)/2.
Hint
Use energy conservation and rolling condition ω = v/R.